Question

已知sin(a+π/6)=3/5，a∈（0，π/2），求sina

Answer 1

sin(a+π/6)=3/5，a∈（0，π/2）
所以
cos(a+π/6)=4/5

所以
sina=sin(a+π/6-π/6)
=sin(a+π/6)cosπ/6-cos(a+π/6)sinπ/6
=3/5*√3/2-(4/5)*1/2
=(3√3-4)/10

Answer 2

sin(a+π/6)=3/5,a∈（0，π/2），所以cos(a+π/6)=4/5
sin(a+π/6-π/6)=sin(a+π/6)cosπ/6-cos(a+π/6)sinπ/6
=3/5乘以根号3/2-4/5乘以1/2
=3根号3/10-2/5

Answer 3

a∈（0，π/2）
a-π/6∈（-π/6，π/3）
sin（a-π/6）=3/5,所以 cos（a-π/6）=4/5
sina=sin[(a-π/6)+π/6]
=sin(a-π/6)sin(π/6)+cos(a-π/6)cos(π/6)
=(3/5)*(1/2)+(4/5)*(√3/2)
=(3+4√3)/10

Answer 4

a∈（0，π/2）
a+π/6∈（π/6，2π/3）

sin(a+π/6)=3/5
cos(a+π/6)=±4/5

sina
=sin[(a+π/6)-π/6]
=sin(a+π/6)cosπ/6-cos(a+π/6)sinπ/6
=3/5*√3/2-(±4/5)*1/2
=(3√3±4/5)/10

Answer 5

cos(a+π/6)=-2/5
Sina=sin〖(a+π/6)-π/6〗=sin(a+π/6)cosπ/6-sinπ/6cos(a+π/6)
=3/5× /2-1/2×(-2/5)=(3 +2)/10